Laplace Transform Methods

The Laplace transform converts linear ODEs with initial conditions into algebra. Learn the definition and existence, standard transform pairs, linearity, first and second shifting theorems, transforms of derivatives and integrals, unit-step and impulse functions, inverse transforms by partial fractions, convolution, and initial/final value theorems.

For circuits switching on, beams under sudden loads or control systems responding to steps, Laplace transforms handle initial conditions and discontinuous inputs automatically — making them the standard tool of electrical and control engineering.

An analogy

Laplace is a translator: take a hard calculus problem into the 's-world' where it becomes algebra, solve it there, and translate back.

Solving IVPs

Transform the ODE (initial conditions enter automatically), solve algebraically for Y(s), decompose into partial fractions, and invert using the table.

Steps and impulses

Switched or delayed inputs use u(t − a) and the second shifting theorem. Impulses (hammer blows, capacitor current spikes) transform to 1 (or e^(−as)).

Transfer functions

With zero initial conditions, Y(s) = G(s)U(s); G(s) captures the system. The impulse response is ℒ⁻¹{G}; outputs for any input are convolutions.

Theorems and pitfalls

Final value theorem valid only if sF(s) has all poles in the left half-plane. Complex-conjugate poles invert to damped sinusoids (complete the square).

Formulas

• Definition: F(s) = ∫₀^∞ f(t)e^(−st) dt
• Pairs: 1 ↔ 1/s, t ↔ 1/s², tⁿ ↔ n!/sⁿ⁺¹, e^(at) ↔ 1/(s − a), sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²)
• Derivatives: ℒ{y′} = sY − y(0); ℒ{y″} = s²Y − sy(0) − y′(0)
• Shifting: ℒ{e^(at)f} = F(s − a); ℒ{f(t − a)u(t − a)} = e^(−as)F(s)
• Convolution: ℒ{f * g} = F(s)G(s)
• Initial / final value: f(0⁺) = lim_{s→∞} sF(s); f(∞) = lim_{s→0} sF(s)

Worked example: Solve an IVP

Given: y″ + 3y′ + 2y = 0, y(0) = 1, y′(0) = 0

Find: y(t).

• s²Y − s + 3(sY − 1) + 2Y = 0 → Y = (s + 3)/((s + 1)(s + 2))
• Partial fractions: 2/(s + 1) − 1/(s + 2)

Answer: y = 2e^(−t) − e^(−2t)

Worked example: Final value

Given: Y(s) = 10/(s(s + 5))

Find: y(∞).

• sY = 10/(s + 5) → s → 0 gives 2

Answer: 2

Common mistakes

• Wrong: Forgetting initial conditions in ℒ{y′}. — Right: ℒ{y′} = sY − y(0).
• Wrong: Final value theorem on unstable systems. — Right: Check poles of sF(s).

Applications

• Circuit transients: Switching RLC networks.
• Control systems: Transfer functions and step responses.
• Structural dynamics: Impulse and step loads.

Key takeaways

• Transform → algebra → partial fractions → invert.
• Initial conditions enter via ℒ{y′}, ℒ{y″}.
• Steps/impulses via shifting theorems.
• FVT only for stable systems.

Going further

• Bromwich integral: Inverse transform via contour integration.
• Two-sided Laplace: Non-causal signals.
• Laplace for PDEs: Transforming time in the heat equation.